1497. Design A Stack With Increment Operation¶

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1497. Design a Stack With Increment Operation

Medium

Design a stack that supports increment operations on its elements.

Implement the CustomStack class:

• CustomStack(int maxSize) Initializes the object with maxSize which is the maximum number of elements in the stack.
• void push(int x) Adds x to the top of the stack if the stack has not reached the maxSize.
• int pop() Pops and returns the top of the stack or -1 if the stack is empty.
• void inc(int k, int val) Increments the bottom k elements of the stack by val. If there are less than k elements in the stack, increment all the elements in the stack.

Example 1:

Input
["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"]
[[3],[1],[2],[],[2],[3],[4],[5,100],[2,100],[],[],[],[]]
Output
[null,null,null,2,null,null,null,null,null,103,202,201,-1]
Explanation
CustomStack stk = new CustomStack(3); // Stack is Empty []
stk.push(1);                          // stack becomes [1]
stk.push(2);                          // stack becomes [1, 2]
stk.pop();                            // return 2 --> Return top of the stack 2, stack becomes [1]
stk.push(2);                          // stack becomes [1, 2]
stk.push(3);                          // stack becomes [1, 2, 3]
stk.push(4);                          // stack still [1, 2, 3], Do not add another elements as size is 4
stk.increment(5, 100);                // stack becomes [101, 102, 103]
stk.increment(2, 100);                // stack becomes [201, 202, 103]
stk.pop();                            // return 103 --> Return top of the stack 103, stack becomes [201, 202]
stk.pop();                            // return 202 --> Return top of the stack 202, stack becomes [201]
stk.pop();                            // return 201 --> Return top of the stack 201, stack becomes []
stk.pop();                            // return -1 --> Stack is empty return -1.

Constraints:

• 1 <= maxSize, x, k <= 1000
• 0 <= val <= 100
• At most 1000 calls will be made to each method of increment, push and pop each separately.

Solution¶

class CustomStack {
    static int max;
    static int arr[];
    static int right;
    public CustomStack(int maxSize) {
        right = 0;
        max = maxSize;
        arr = new int[max];
    }

    public void push(int x) {
        if (right == max) return;
        arr[right] = x;
        right++;
    }

    public int pop() {
        if (right == 0) return -1;
        int ele = arr[right - 1];
        right--;
        return ele;
    }

    public void increment(int k, int val) {
        for (int i = 0; i < Math.min(k , right); i++) arr[i] += val;
    }
}

/**
 * Your CustomStack object will be instantiated and called as such:
 * CustomStack obj = new CustomStack(maxSize);
 * obj.push(x);
 * int param_2 = obj.pop();
 * obj.increment(k,val);
 */

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here